What would be the way to select elements when two conditions are True in a matrix?
In R, it is basically possible to combine vectors of booleans.
So what I'm aiming for:
A = np.array([2,2,2,2,2])
A < 3 and A > 1 # A < 3 & A > 1 does not work either
Evals to:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
It should eval to:
array([True,True,True,True,True])
My workaround usually is to sum these boolean vectors and equate to 2, but there must be a better way. What is it?
you could just use &, eg:
x = np.arange(10)
(x<8) & (x>2)
gives
array([False, False, False, True, True, True, True, True, False, False], dtype=bool)
A few details:
- This works because
& is shorthand for the numpy ufunc bitwise_and, which for the bool type is the same as logical_and. That is, this could also be spelled out as
bitwise_and(less(x,8), greater(x,2))
- You need the parentheses because in numpy
& has higher precedence than < and >
and does not work because it is ambiguous for numpy arrays, so rather than guess, numpy raise the exception.
There's a function for that:
In [8]: np.logical_and(A < 3, A > 1)
Out[8]: array([ True, True, True, True, True], dtype=bool)
Since you can't override the and operator in Python it always tries to cast its arguments to bool. That's why the code you have gives an error.
Numpy has defined the __and__ function for arrays which overrides the & operator. That's what the other answer is using.
While this is primitive, what is wrong with
A = [2, 2, 2, 2, 2]
b = []
for i in A:
b.append(A[i]>1 and A[i]<3)
print b
The output is [True, True, True, True, True]
